STRUCTURE OF I-127
.By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. .NUCLEAR STRUCTURE OF STABLE IODINE (I-127) WITH 29 BLANK POSITIONS There are 37 known isotopes of iodine (I) from I-108 to I-144, but only one, I-127, is stable. Iodine is thus a monoisotopic element. Comparing the stable Iodine of 53 protons (odd number) with the Tellurium of 52 protons (even number) we conclude that the structure of the stable Iodine brakes the high symmetry of the structure of Tellurium. (See my STRUCTURE OF Te-120...Te-130 ). Note that here there is not any additional proton (p54) to fill the symmetrical blank position of 1(p) existing between n30 and n48. So this situation changes the shape of high symmetry having two squares over and under the six horizontal planes. In the following diagram of I-127 the additional p53n53 is not shown . However the position of p53 of the vertical p53n53 is shown by using the top view of the third horizontal plane. For the changing of the shape of Tellurium the p37n37 and the p38n38 of the two squares are moved to fill the blank positions with S = +2 near the n29p29 and n32p32 . Note that these new arrangements in the fifth horizontal plane reduce the number of 4n to 2n . For example in the fifth plane the n37 replaces the 1n. However the new p37 and p38 o the fifth plane give 2n at the sixth horizontal plane formed by p37 and p31 and p38 and p32 respectively. Also using the top view of the third plane we see that these two planes give 7(n). Under this condition the number N of blank positions is given by The p39n39 and p40n40 give 6n of opposite spins The first plane gives 2(n) of positive spins The sixth plane gives 2(n) and 2n of negative spins The second plane gives 2{n} + 4n of negative spins The fifth plane gives 2{n} + 2n of positive spins The third plane gives 4(n) of positive spins The fourth plane gives 3(n) of negative spins That is N = 6n +2(n) +2(n) + 2n + 2{n} + 4n + 2{n} + 2n +4(n) +3(n) = 29 blank positions. Since the I-127 of 21 extra neutrons has S= +5/2 we conclude that it is due to the S=+2 of the deuterons p37n37 and p38n38 by adding 11 extra neutrons of positive spins and 10 extra neutrons of negative spins. That is S = +2 + 2{+1/2} +3+1/2 +2+1/2 +4(+1/2) +2{-1/2} + 2-1/2 +4-1/2 +2(-1/2) = +5/2 . DIAGRAM OF I-127 FORMING 29 BLANK POSITIONS Here the additional p53n53 is not shown, but you can see the position of p53 by using the top view of the third horizontal plane.. Here you see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Here you see the p37n37 near p29n29 and the p38n38 near the p30n30 which give 1n and 1n near p31 and p32. n40............p40 .............n n40p40 with n n31………p12.........n12.......p32............n n........p31........n11.........p11…… n32 Sixth H. plane n37........ p29.........n10.........p10…… n30.........p38 p37......n29…......p9..........n9 …….p30..........n38 Fifth H. plane p47.......n27.........p8..........n8.........p28......... n48 n45..........p27....n7..............p7........n28..........p46 Fourth H. plane n47......p25.........n6.........p6..........n26...........p48 p45.......n25….. p5..........n5……….p26.....n46 Third H. plane n23………p4........n4………….p24.................n n......p23… ...n3……….p3………..n24 Second H. plane p21.........n2………p2............n22 n21........p1..........n1.........p22 First H. plane n39.........p39.........n n39p39 with n TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 HERE YOU CAN SEE ALSO THE 1n FORMED BY p39 AND p22 (n)........p34....... n34 p21....... n2........ p2....... n22 n21.........p1. .......n1.......p22 n33.......p33..... (n) ' TOP VIEW OF THE SECOND HORIZONTAL PLANE Here the n near the p14 fills the blank position formed by p51 and p14. While the {n} near the p14 fills the blank position formed by p14, p24 and p44. The 2n near p24 and p23 fill the blank positions formed by p24 and p48 as well as by p23 and p45. Moreover the blank position of {n} near p23 is formed by p23, p13 and p41. Finally the blank position of n near the p13 is formed byp13 and p49. That is we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. n n14.......p14........{n} n23.......p4.........n4........p24..........n n.......p23........n3........p3.........n24 {n}...... p13.......n13 TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS Here the p53, np4, n55, p56 and n57 fill the blank positions of (n) and (p). So you can see the existing blank positions for receiving 2(n).and 2(p) n50.......p51......(n) p53........n42........p16......n16......p44.........(n) n47........p25........n6........p6........n26.........p48 p45........n25........p5........n5........p26........ n46 (n)..........p41.......n15.......p15.......n43 (n)........p49.......n52 Category:Fundamental physics concepts